您如何检查给定的句子在Java中是否回避某个字母？

Question

I have been trying to figure out how to see if a given sentence is a lipogram avoiding the letter E. I have gotten to the point where I put in the true/false statements, but it is only outputting "This sentence is not a Lipogram avoiding the letter e. " every time, whether the input contains e or not. What am I doing wrong?

boolean avoidsE = false, hasE = true, avoidsS = false, containsS = true;

for(int i = 0; i < sentence.length(); i++)
{
  if (sentence.charAt(i) == 'e')
  hasE = true;
  else
  avoidsE = false;
}

if (hasE = true)
System.out.println("This sentence is not a Lipogram avoiding the letter e. ");
else if (avoidsE = false)
System.out.println("This sentence is a Lipogram avoiding the letter e! ");

Answer 1

if (hasE == true) // "=" is to assign, you want to use "==" here to check  
    System.out.println("This sentence is not a Lipogram avoiding the letter e. "); 
else if (avoidsE == false) //same here   
    System.out.println("This sentence is a Lipogram avoiding the letter e! ");

Answer 2

If you want to compare something you should use double equals ==

For search if an character is in the string you can use contains method and to check if letter is not contained in the string you can use indexOf . It would be something like this:

public static void main(String[] args) {
    String sentence = "This is your sentence";

    if(sentence.contains("e")){
        System.out.println("This sentence is not a Lipogram avoiding the letter e. ");
    }
    else if("e".indexOf(sentence) < 0){
        System.out.println("This sentence is a Lipogram avoiding the letter e!");
    }
}

Answer 3

No need to reinvent the wheel here.

public class Main {
    public static void main(String[] args) {
        System.out.println(hasChar("asdfe", 'e'));
        System.out.println(hasChar("asdfghij", 'e'));
    }

    static boolean hasChar(String str, char c) {
        return str.chars().anyMatch(x -> x == c);
    }
}

Output:

true
false